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Calculus 2 : Applications in Physics

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Initial Conditions

Find the work done by gravity exerting an acceleration of $-10\frac{m}{s^2}$ for a 10kg block down from its original position with no initial velocity.

Remember that

$F_{grav}=mass*acceleration$

$W=\int_{a}^{b}F(x)dx$ , where F(x) is a force measured in Newtons , is work measured in Joules , and and are initial and final positions respectively.

Possible Answers:

1000J

-500J

100J

250J

Correct answer:

250J

Explanation:

The force of gravity is proportional to the mass of the object and acceleration of the object.

$F=ma= (10*-10)\frac{kg*m}{s^2}= -100N$

Since the block fell down 5 meters, its final position is and initial position is .

$\\ W=\int_{0}^{-5}-100dx\\ \\=\frac{-100x}{2}|_{0}^{-5} \\ \\=-50x|_{0}^{-5}\\ \\=-50(-5)--50(0)\\ \\=250J$

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Example Question #1931 : Calculus Ii

The velocity of a car is defined by the equation $v(t)=4x^{2}-3x$ , where is the time in minutes. What distance (in meters) does the car travel between t=0 and t=4 ?

Possible Answers:

600m

60 m

$\frac{3}{184} m$

$\frac{184}{3} m$

$\frac{190}{3}m$

Correct answer:

$\frac{184}{3} m$

Explanation:

We define velocity as the derivative of distance, or v(t)=d'(t) .

Thus, in order to find the distance traveled when given the velocity, we'll need to take the definite integral of the velocity function over a specified time period, or $\int_{a}^{b}v(t)dt$ .

Since $v(t)=4x^{2}-3x$ , we can use the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

to find:

$\int_{0}^{4}(4x^{2}-3x)dx$

$=\left[\frac{4}{3}x^{3}-\frac{3}{2}x^{2}\right]_{0}^{4}\textrm{}$

$=\left[\frac{4}{3}(4)^{3}-\frac{3}{2}(4)^{2}\right]-\left[\frac{4}{3}(0)^{3}-\frac{3}{2}(0)^{2}\right]$

$=\left[\frac{256}{3}-\frac{48}{2}\right]$

$=\frac{512}{6}-\frac{144}{6}$

$=\frac{368}{6}$

$=\frac{184}{3}$

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Example Question #11 : Integral Applications

The velocity of a train is defined by the equation $v(t)=x^{2}+6x-5$ , where is the time in seconds What distance (in meters) does the train travel between t=0 and t=3 ?

Possible Answers:

20m

21 m

10m

12m

25m

Correct answer:

21 m

Explanation:

We define velocity as the derivative of distance, or v(t)=d'(t) .

Thus, in order to find the distance traveled when given the velocity, we'll need to take the definite integral of the velocity function over a specified time period, or $\int_{a}^{b}v(t)dt$ .

Since $v(t)=x^{2}+6x-5$ , we can use the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

to find:

$\int_{0}^{3}(x^{2}+6x-5)dx$

$=\left[\frac{1}{3}x^{3}+3x^{2}-5x\right]_{0}^{3}\textrm{}$

$=\left[\frac{1}{3}(3)^{3}+3(3)^{2}-5(3)\right]-\left[\frac{1}{3}(0)^{3}+3(0)^{2}-5(0)\right]$

$=\frac{1}{3}(3)^{3}+3(3)^{2}-5(3)$

$=\frac{27}{3}+27-15$

=9+27-15

=36-15

=21

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Example Question #13 : Integral Applications

The velocity of a balloon is defined by the equation $v(t)=6x^{2}-4x+12$ , where is the time in minutes. What distance (in meters) does the balloon travel between t=0 and t=2 ?

Possible Answers:

34m

32m

30m

23m

33m

Correct answer:

32m

Explanation:

We define velocity as the derivative of distance, or v(t)=d'(t) .

Thus, in order to find the distance traveled when given the velocity, we'll need to take the definite integral of the velocity function over a specified time period, or $\int_{a}^{b}v(t)dt$ .

Since $v(t)=6x^{2}-4x+12$ , we can use the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

to find:

$\int_{0}^{2}(6x^{2}-4x+12)dx$

$=[2x^{3}-2x^{2}+12x]_{0}^{2}\textrm{}$

$=[2(2)^{3}-2(2)^{2}+12(2)]-[2(0)^{3}-2x=(0)^{2}+12(0)]$

=16-8+24

=8+24

=32

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Example Question #14 : Integral Applications

A frisbee has a velocity defined by v(t)=2t+13 , where we express in seconds. What distance does it travel between $0\leq t\leq 2$ in meters?

Possible Answers:

30m

32m

34m

28m

26m

Correct answer:

30m

Explanation:

We define velocity as the derivative of distance, or v(t)=d'(t) .

Thus, in order to find the distance traveled when given the velocity, we'll need to take the definite integral of the velocity function over a specified time period, or

$\int_{a}^{b}v(t)dt$ .

Since

v(t)=2t+13 ,

we can use the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

to find:

$\int_{0}^{2}(2t+13)dt$

$=[t^{2}+13t]_{0}^{2}\textrm{}$

$=[(2)^{2}+13(2)]-[(0)^{2}+(0)t]$

=4+26

=30

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Example Question #15 : Integral Applications

A ball has a velocity defined by v(t)=3t-4 , where we express in seconds. What distance does it travel between $0\leq t\leq 3$ in meters?

Possible Answers:

$\frac{5}{2}m$

$\frac{5}{3}m$

$\frac{2}{3}m$

None of the above

$\frac{3}{2}m$

Correct answer:

$\frac{3}{2}m$

Explanation:

We define velocity as the derivative of distance, or v(t)=d'(t) .

Thus, in order to find the distance traveled when given the velocity, we'll need to take the definite integral of the velocity function over a specified time period, or $\int_{a}^{b}v(t)dt$ .

Since

v(t)=3t-4 , we can use the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

to find:

$\int_{0}^{3}(3t-4)dt$

$=[\frac{3}{2}t^{2}-4t]_{0}^{3}$

$=[\frac{3}{2}(3)^{2}-4(3)]-[\frac{3}{2}(0)^{2}-4(0)]$

$=\frac{27}{2}-12$

$=\frac{27}{2}-\frac{24}{2}$

$=\frac{3}{2}$

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Example Question #11 : Applications In Physics

A subway has a velocity defined by v(t)=11t+9 , where we express in seconds. What distance does it travel between $0\leq t\leq 4$ in meters?

Possible Answers:

128m

126m

124m

130m

122m

Correct answer:

124m

Explanation:

We define velocity as the derivative of distance, or v(t)=d'(t) .

Thus, in order to find the distance traveled when given the velocity, we'll need to take the definite integral of the velocity function over a specified time period, or $\int_{a}^{b}v(t)dt$ .

Since

v(t)=11t+9 ,

we can use the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

to find:

$\int_{0}^{4}(11t+9)dt$

$=[\frac{11}{2}t^{2}+9t]_{0}^{4}\textrm{}$

$=\frac{176}{2}+36$

=88+36

=124

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Example Question #17 : Integral Applications

A train goes a certain distance between $0\leq t \leq 2$ (where is time in seconds). If we know that the train's velocity is defined as v(t)=12t+15 , what is the distance it travelled (in meters)?

Possible Answers:

56m

50m

45m

46m

54m

Correct answer:

54m

Explanation:

We define velocity as the derivative of distance, or v(t)=d'(t) .

Thus, in order to find the distance traveled when given the velocity, we'll need to take the definite integral of the velocity function over a specified time period, or $\int_{a}^{b}v(t)dt$ .

Since v(t)=12t+15 , we can use the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

to find:

$\int_{0}^{2}(12t+15)dt$

$=[6t^{2}+15t]_{0}^{2}\textrm{}$

$=[6(2)^{2}+15(2)]-[6(0)^{2}+15(0)]$

=[24+30]

=54

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Example Question #18 : Integral Applications

A car goes a certain distance between $0\leq t \leq 3$ (where is time in seconds). If we know that the car's velocity is defined as v(t)=14t-2 , what is the distance it travelled (in meters)?

Possible Answers:

55m

59m

58m

56m

57m

Correct answer:

57m

Explanation:

We define velocity as the derivative of distance, or v(t)=d'(t) .

Thus, in order to find the distance traveled when given the velocity, we'll need to take the definite integral of the velocity function over a specified time period, or $\int_{a}^{b}v(t)dt$ .

Since v(t)=14t-2 , we can use the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

to find:

$\int_{0}^{3}(14t-2)dt$

$=[7t^{2}-2t]_{0}^{3}\textrm{}$

$=[7(3)^{2}-2(3)]-[7(0)^{2}-2(0)]$

=[63-6]

=57

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Example Question #11 : Integral Applications

A plane goes a certain distance between $2\leq t \leq 5$ (where is time in seconds). If we know that the plane's velocity is defined as v(t)=6t-2 , what is the distance it travelled (in meters)?

Possible Answers:

60m

61m

57m

58m

59m

Correct answer:

57m

Explanation:

We define velocity as the derivative of distance, or v(t)=d'(t) .

Thus, in order to find the distance traveled when given the velocity, we'll need to take the definite integral of the velocity function over a specified time period, or $\int_{a}^{b}v(t)dt$ .

Since v(t)=6t-2 , we can use the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

to find:

$\int_{2}^{5}(6t-2)dt$

$=[3t^{2}-2t]_{2}^{5}\textrm{}$

$=[3(5)^{2}-2(5)]-[3(2)^{2}-2(2)]$

=75-10-12+4

=57

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Calculus 2 : Applications in Physics

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Initial Conditions

Example Question #1931 : Calculus Ii

Example Question #11 : Integral Applications

Example Question #13 : Integral Applications

Example Question #14 : Integral Applications

Example Question #15 : Integral Applications

Example Question #11 : Applications In Physics

Example Question #17 : Integral Applications

Example Question #18 : Integral Applications

Example Question #11 : Integral Applications

All Calculus 2 Resources

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